If Ag atoms occupy corners of a cube in a lattice and Cu occupies the centers, what's the simplest formula?
Answer:
The simplest formula is AgCu3.
The one you described is an example of a face centered cubic which has 8 atoms on the corners (1/8 contribution for each corner) and 4 atoms on each face (1/2 contribution for each face). Since there are 8 corners, 1/8 * 8 equals 1; and since there are 6 faces in a cube, 1/2 * 6 equals 3.
simplest formula is AgCu
8 corners of the cubic lattice each contricute 1/8th Ag to the unit cell (u have to 'see' this by using ur 3d visualization and imagination), so total in one cube there are 1/8 * 8 = 1 Ag atom,
plus 1 Cu atom at the centre.
formula is AgCu
No. In one cube, the 8 Ag atoms are not completely in. In fact, each Ag atom is only 1/8 in the cube, so there is 8*(1/8)=1 Ag atoms per cube.
It is a bit ambiguous whether the Cu are at the centers of the cubes or the centers of the faces. For the centers of the cube, you will get AgCu.
If the Cu are at the centers of the faces, there are 6 Cu atoms, but each one is only 1/2 in the cube, so that give 6*(1/2)=3 Cu atoms per cube, for a formula of AgCu_3.
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